问题及解答

设 $z=f(xy^2,\ln(xy))$, 求 $\dfrac{\partial^2 z}{\partial x^2}$, $\dfrac{\partial^2 z}{\partial y^2}$, $\dfrac{\partial^2 z}{\partial x\partial y}$.

\[
\frac{\partial z}{\partial x}=f'_1\cdot y^2+f'_2\cdot\frac{1}{xy}\cdot y=y^2 f'_1+\frac{1}{x}f'_2\ ,
\]

\[
\frac{\partial z}{\partial y}=f'_1\cdot 2xy+f'_2\cdot\frac{1}{xy}\cdot x=2xy f'_1+\frac{1}{y}f'_2\ ,
\]

\[
\begin{split}
\frac{\partial^2 z}{\partial x^2}&=\frac{\partial}{\partial x}\biggl(\frac{\partial z}{\partial x}\biggr)=\frac{\partial}{\partial x}\Bigl(y^2 f'_1+\frac{1}{x}f'_2\Bigr)=y^2\cdot\Bigl(f''_{11}y^2+f''_{12}\frac{1}{xy}\cdot y\Bigr)+\frac{-1}{x^2}f'_2+\frac{1}{x}\Bigl(f''_{21}y^2+f''_{22}\frac{1}{xy}\cdot y\Bigr)\\
&=y^4f''_{11}+\frac{y^2}{x}f''_{12}-\frac{1}{x^2}f'_2+\frac{y^2}{x}f''_{21}+\frac{1}{x^2}f''_{22}\ ,
\end{split}
\]

\[
\begin{split}
\frac{\partial^2 z}{\partial y^2}&=\frac{\partial}{\partial y}\biggl(\frac{\partial z}{\partial y}\biggr)=\frac{\partial}{\partial y}\Bigl(2xyf'_1+\frac{1}{y}f'_2\Bigr)=\Bigl[2xf'_1+2xy(f''_{11}\cdot 2xy+f''_{12}\cdot\frac{1}{xy}\cdot x)\Bigr]+\Bigl[\frac{-1}{y^2}f'_2+\frac{1}{y}(f''_{21}\cdot 2xy+f''_{22}\cdot\frac{1}{xy}\cdot x)\Bigr]\\
&=\Bigl[2xf'_1+4x^2 y^2 f''_{11}+2xf''_{12}\Bigr]+\Bigl[-\frac{1}{y^2}f'_2+2xf''_{21}+\frac{1}{y^2}f''_{22}\Bigr]\\
&=2xf'_1-\frac{1}{y^2}f'_2+4x^2 y^2 f''_{11}+4xf''_{12}+\frac{1}{y^2}f''_{22}\ ,
\end{split}
\]

\[
\begin{split}
\frac{\partial^2 z}{\partial x\partial y}&=\frac{\partial}{\partial y}\biggl(\frac{\partial z}{\partial x}\biggr)=\frac{\partial}{\partial y}\Bigl(y^2f'_{1}+\frac{1}{x}f'_2\Bigr)\\
&=2yf'_1+y^2(f''_{11}\cdot 2xy+f''_{12}\cdot\frac{1}{xy}\cdot x)+\frac{1}{x}(f''_{21}\cdot 2xy+f''_{22}\cdot\frac{1}{xy}\cdot x)\\
&=2yf'_1+2xy^3 f''_{11}+yf''_{12}+2yf''_{21}+\frac{1}{xy}f''_{22}\ ,
\end{split}
\]

注意这里 $f\in C^2$, 故 $f''_{12}=f''_{21}$, 故还可进一步简化.